# What is the derivative of this function y=tan^-1(2x^4)?

May 30, 2017

$\frac{8 {x}^{3}}{4 {x}^{8} + 1}$

#### Explanation:

Aside from just applying the chain rule, one helpful way to perform this derivation may be to substitute a variable u to break the equation into separate pieces of a puzzle, like this:

We know the chain rule is
$\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dx}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

We can also let u (in this case) = $2 {x}^{4}$

Now just take the derivative of the function in two puzzle pieces:

We know $\frac{d}{\mathrm{du}}$ ${\tan}^{-} 1 \left(u\right)$ is just $\frac{d}{\mathrm{dx}} \left(T a {n}^{-} 1 \left(x\right)\right) = \frac{1}{{x}^{2} + 1}$

SO...

$\frac{d}{\mathrm{du}} \left({\tan}^{-} 1 \left(u\right)\right) = \frac{1}{{u}^{2} + 1}$

and

$\frac{d}{\mathrm{dx}} \left(2 {x}^{4}\right) = 8 {x}^{3}$ (By the power rule)

From here, just substitute back $u = 2 {x}^{4}$ to get $\frac{1}{{\left(2 {x}^{4}\right)}^{2} + 1} \cdot 8 {x}^{3}$

Which simplifies to... $\frac{8 {x}^{3}}{4 {x}^{8} + 1}$

V'oila! Our puzzle is complete!